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2u Mathematics Marathon v1.0 (1 Viewer)

SoulSearcher

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drynxz said:
ill take a shot even thou i havent learnt it yet...

2x/(x-2) >1

(multiply both sides by (x-2)^2)
= 2x(x-2) > (x-2)^2
= 2x^2 - 4x >x^2-4x+4
= x^2 - 4 >0
= (x+2)(x-2)>0
*test point x=0
= (2)(-2)>0 doesnt hold true
therefore
= x > 2
pft maybe i got it wrong...i dunno
In the bold lines, the first 2 lines are correct, but you forgot to account for the 2 values of x. Therefore the correct answer is x < -2 or x > 2
 

SoulSearcher

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drynxz said:
Next question:



Solve ==> solve simultaneously



x+2y-z = -5

2x-3y+4z = 28

4x+5y-3z = -10


some substitution is involved here

x + 2y - z = -5 ... (1), thus z = x + 2y + 5
2x-3y+4z = 28 ... (2)
4x+5y-3z = -10 ... (3)

sub in the value of z into (2) & (3)

2x - 3y + 4(x + 2y + 5) = 28, thus
6x + 5y = 8 ... (4)
4x + 5y - 3(x + 2y + 5) = -10, thus
x - y = 5 ... (5)
{ (5) x 5 } 5x - 5y = 25 ... (6)

add (6) to (4)

11x = 33
x = 3

sub value of x into (5)

3 - y = 5
y = -2

sub values of x and y into equation for z

z = 3 - 4 + 5
z = 4

therefore x = 3, y = -2 and z = 4

Next Question:

A bowl is formed by rotating the part of the curve y = x4 / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.
 
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Mountain.Dew

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SoulSearcher said:
Next Question:

A bowl is formed by rotating the part of the curve y = x4 / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.
y = x4 / 4, so 4y = x4, x = (4y)1/4

when x = 0, y = 0.
when x= 2, y = 4.

so ur volume is thus:

V = pi * (Integral from 0 to 4 of (4y)1/2)dy

V = pi * [(4y)3/2 / 6], 0 --> 4

V = pi * 64/6 = 32/3pi units3

hopefully that should be right.



Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.
 
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drynxz

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Next Question:

A bowl is formed by rotating the part of the curve y = x4 / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.
can someone tell me what i did wrong for soulsearchers question :S

V= *intergral B over A pi y 2 dx

=*intergral 2 over 0 pi (x4/4)2 dx

=*intergral 2 over 0 pi x8/16 dx

=pi [x^9/144x]2 over 0

=pi [512/288]

=(16/9) pi

=5.585 units 3
yeh it seem wrong..lol maybe i used the wrong formula or some crap :S
 

Mountain.Dew

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drynxz said:
can someone tell me what i did wrong for soulsearchers question :S

V= *intergral B over A pi y 2 dx

=*intergral 2 over 0 pi (x4/4)2 dx

=*intergral 2 over 0 pi x8/16 dx

=pi [x^9/144x]2 over 0

=pi [512/288]

=(16/9) pi

=5.585 units 3
yeh it seem wrong..lol maybe i used the wrong formula or some crap :S
mmmmmm theres something wrong here...

remember the volume is made by rotating the curve around the Y-AXIS not the x-axis, so you need to get y the subject from the original curve. you also needed to change the limits as well.

please see my solution (previous post).

oh, and with powers, it is (suP)...(/suP), not (suB)...(/suB) (replace the ( ) with [ ] though)
 

drynxz

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yeh i realised i read the question wrong...and i guess its pretty obvious im new to this thread -_- lol
 

sando

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has anyone got a question for me? preferably integration, cos i hav a test on that soon
 

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pLuvia said:
Ok try this

∫ sec3x tanx dx
∫ sec3x tanx dx = ∫ sec2x dx = tanx

I'm just guessing that works, not 100% sure.
 
P

pLuvia

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I'm not sure how you got that PLooB but here's the solution

∫ sec3x tanx dx

Let u = secx
du = secxtanx dx

.: ∫ sec3x tanx dx
= ∫ sec2x (secxtanx dx)
= ∫ u2 du
= u3 / 3 + C
 

Mountain.Dew

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pLuvia said:
I'm not sure how you got that PLooB but here's the solution

∫ sec3x tanx dx

Let u = secx
du = secxtanx dx

.: ∫ sec3x tanx dx
= ∫ sec2x (secxtanx dx)
= ∫ u2 du
= u3 / 3 + C
make sure u sub u = secx back into ur primtive function to get the correct answer ;)

heres another question (which i posted earlier, lost in the sea of posts)

Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.
 

insert-username

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Mountain.Dew said:
Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.
Annuities is 2U, but this one is particularly hard, since it's not straightforward and requires fiddling with the resulting equation. You might want to try an easier one to keep it 2U. :)


I_F
 

Mountain.Dew

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insert-username said:
Annuities is 2U, but this one is particularly hard, since it's not straightforward and requires fiddling with the resulting equation. You might want to try an easier one to keep it 2U. :)


I_F
oh dear. sorry about that guys. i will post up a new question that is 2U.

in the meantime, the 3U question is still up there for your interest and viewing pleasure, as the saying goes.

Next Question:

a particle travels in a straight line. its position x cm from the centre at time t is given by the equation: x = 6 - 5t - 2t^2

a) find the initial velocity
b) find when the particle will come to rest
c) find time when the particle passes through the centre.
 

SoulSearcher

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Mountain.Dew said:
Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.
This question is just an extension of the 2 unit course, but it isn't only for 3 unit students, it just requires some index law fiddling.

Now
An = 24000(1.15)n - { 3840 ( 1.15n - 1) } / { 1.15 - 1 }

but at the end of the loan, An = 0, therefore

0 = 24000(1.15)n - { 3840 ( 1.15n - 1) } / { 1.15 - 1 }
0 = 24000(1.15)n - 25600( 1.15n - 1)
25600( 1.15n - 1) = 24000(1.15)n
25600( 1.15n ) - 24000( 1.15n ) = 25600
1600( 1.15n ) = 25600
1.15n = 16

now remembering 1.15n = 1 / 1.15-n
1 / 1.15-n = 16
1 = 16(1.15-n)
0.0625 = 1.15-n

add 1 to both sides
1 + 0.0625 = 1.15-n + 1

take the values that are not equal to 1 over
1 - 1.15-n = 0.9375

thus showing what was needed.

to find n , refer back to the working out
1.15n = 16
n ln 1.15 = ln 16
n = ln 16 / ln 1.15
n = 19.83
n=20 rounded off to nearest integer (edited)

Ill do the other question later
 
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Mountain.Dew

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SoulSearcher, remember n is an integer, because u it involves the element of reality.

therefore, n = 20 installments.
 

SoulSearcher

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Whoops, edited now

Mountain.Dew said:
Next Question:

a particle travels in a straight line. its position x cm from the centre at time t is given by the equation: x = 6 - 5t - 2t^2

a) find the initial velocity
b) find when the particle will come to rest
c) find time when the particle passes through the centre.
(a) x = 6 - 5t - 2t2
therefore dx/dt = -5 - 4t which is velocity
when t = 0,
v = -5 - 0
therefore initial velocity is -5cms-1

(b) to find the time when the point rests, v = 0
0 = -5 - 4t
4t = -5
t = -5/4 (notice here that time cannot be negative and therefore there is no resting point)

(c)this occurs when x = 0
0 = 6 - 5t - 2t2
2t2 + 5t - 6 = 0
since that cannot be factorised easily, use quadratic formula
t = { -5 ± sqrt(25 + 48) } / 4
t = { -5 ± sqrt(73) } / 4
therefore t = ( -5 - sqrt(73) } / 4 or t = { -5 + sqrt(73) }/4
but as time cannot be negative, t = { -5 + sqrt(73) }/4
 
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Sober

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Mountain.Dew said:
time to revive an old thread...

Question:
y = loge(tanx) + x^2 + 25x - 12

find y'
y' = sec²(x) / tan(x) + 2x + 25

= sec(x).cosec(x) + 2x + 25

Early bird gets to post the next question.
 
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Question continued:
Hence find ∫ 3.sec(x).cosec(x) + sin(5-2x) dx
 

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3ln(tan(x)) + cos(5-2x)/2

y = sqrt(x*tan(x)), find dy/dx

Edit: Fixed -- thanks Riviet :)
 
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